Check variable equals a certain number - php

Question

I need to check to make sure a variable equals either 5, 10, 50 or 100. If it doesn't then I want to set it to 5. Can someone tell me what's wrong with this IF statement?
if (isset($_REQUEST['number']) && $_REQUEST['number'] !== "5" || $_REQUEST['number'] !== "10" || $_REQUEST['number'] !== "50" || $_REQUEST['number'] !== "100") {
$number = 5;
} else {
$number = $_REQUEST['number'];
}

Answer 1

You need to break up your logic to overcome operator precedence. When in doubt, use parenthesis.
I also inverted the logic. It reads more naturally.
if (isset($_REQUEST['number']) && ($_REQUEST['number'] == "5" || $_REQUEST['number'] == "10" || $_REQUEST['number'] == "50" || $_REQUEST['number'] == "100")) {
  $number = $_REQUEST['number'];
} else {
  $number = 5;
}
While the above works, you could also streamline it with something like in_array().
isset($_REQUEST['number']) && in_array($_REQUEST['number'], array(5, 10, 50, 100))

Answer 2

Try:
if (isset($_REQUEST['number']) && ($_REQUEST['number'] == "5" || $_REQUEST['number'] == "10" || $_REQUEST['number'] == "50" || $_REQUEST['number'] == "100")) {
$number = $_REQUEST['number'];
} else {
$number = 5;
}

Answer 3

You've got some very confusing code. Does nothing you said you wanted.
this should be it:
if (isset($_REQUEST['number']) && in_array($_REQUEST['number'], array(5, 10, 50, 100))) {
$number = (int) $_REQUEST['number'];
} else {
$number = 5;
}
Combined from other answers so thank everyone :)

Answer 4

You need to swap things around a bit:
if (!isset($_REQUEST['number']) || ($_REQUEST['number'] !== "5" && $_REQUEST['number'] !== "10" && $_REQUEST['number'] !== "50" && $_REQUEST['number'] !== "100")) {
$number = 5;
} else {
$number = $_REQUEST['number'];
}
Narration of the above: if number isn't set, make it 5. If it is set, but isn't 5, 10, 50, or 100, set it to 5. Otherwise, leave it as-is.

Answer 5

It may be easier in the long run to break the logic up into a different format to try to be a bit clearer, rather then having a single 'if' statement.
$acceptable = array(5, 10, 50, 100);
$number = null;
if (isset($_REQUEST['number']) {
$number = intval($_REQUEST['number']);
}
if (!in_array($number, $acceptable)) {
$number = 5;
}
But with keeping it as a single if statement:
if (
isset($_REQUEST['number']) && (
$_REQUEST['number'] === "5" ||
$_REQUEST['number'] === "10" ||
$_REQUEST['number'] === "50" ||
$_REQUEST['number'] === "100"
)
) {
$number = $_REQUEST['number'];
} else {
$number = 5;
}

Answer 6

Although iMoses, and others already laid down the basics of simplifying your routine, I would suggest making the in_array() check more strict. If you don't and $_REQUEST['number'] would, for instance equal '10F', it would also pass.
// I like to define default values up front.
$number = 5;
// since $_REQUEST['number'] should always be a string
// make in_array() check strictly against these string values
$allowedNumbers = array( '5', '10', '50', '100' );
// make in_array check strictly
if( isset( $_REQUEST['number'] ) && in_array( $_REQUEST['number'], $allowedNumbers, true ) )
{
// get proper integer value
$number = intval( $_REQUEST['number'] );
}
var_dump( $number );